Chapter 5.5, Problem 27E

(a)

To determine

To find : The trapezoidal rule approximations for $n=10,100,1000$ .

Answer to Problem 27E

The answer:

  $\begin{array}{c}{T}_{10}\approx 1.983523538\\ T_{100}\approx 1.999835504\\ T_{1000}\approx 1.999998355\end{array}$

Explanation of Solution

Given information:

The integral is ${{\displaystyle \int }}_0^{\pi }\sin xdx$

Calculation:

To use the calculator program and get,

  $\begin{array}{c}{T}_{10}\approx 1.983523538\\ T_{100}\approx 1.999835504\\ T_{1000}\approx 1.999998355\end{array}$

Therefore, the required trapezoidal rule approximations for $n=10,100,1000$ are:

  $\begin{array}{c}{T}_{10}\approx 1.983523538\\ T_{100}\approx 1.999835504\\ T_{1000}\approx 1.999998355\end{array}$

(b)

To determine

To find: The errors with as many decimal places with accuracy.

Answer to Problem 27E

The answer is:

$n$	$\left|{\text{E}}_T\right|=2-{\text{T}}_n$
10	$1.6476462\times {10}^{-2}$
100	$1.64496\times {10}^{-4}$
1000	$1.645\times {10}^{-6}$

Explanation of Solution

Given information:

The integral is ${{\displaystyle \int }}_0^{\pi }\sin xdx$

Calculation:

From part $(a)$ , the trapezoidal approximations are:

  $\begin{array}{c}{T}_{10}\approx 1.983523538\\ T_{100}\approx 1.999835504\\ T_{1000}\approx 1.999998355\end{array}$

The exact value is:

  $\begin{array}{c}{{\displaystyle \int }}_0^{\pi }\sin xdx=-{\left.\cos x\right|}_0^{\pi }\\ =-\cos \pi +\cos 0\\ =-(-1)+1\\ =2\end{array}$

The errors are then,

$n$	$\left|{\text{E}}_T\right|=2-{\text{T}}_n$
10	$1.6476462\times {10}^{-2}$
100	$1.64496\times {10}^{-4}$
1000	$1.645\times {10}^{-6}$

(c)

To determine

To find: The pattern.

Answer to Problem 27E

The answer is:

  $\left|E_{T_{10n}}\right|\approx {10}^{-2}\left|E_{T_n}\right|$

Explanation of Solution

Given information:

The integral is ${{\displaystyle \int }}_0^{\pi }\sin xdx$

Calculation:

From part $(b)$ , the errors were approximately,

  $\begin{array}{c}\left|E_{T_{10}0}\right|\approx 1.65\times {10}^{-2},\\ \left|E_{T_{100}}\right|\approx 1.65\times {10}^{-4}\text{, }\\ \left|E_{T_{1000}}\right|\approx 1.65\times {10}^{-6}\end{array}$

The error for $T_{100}$ is then about ${10}^{-2}$ times the error for $T_{10}$ and the error for $T_{1000}$ is about ${10}^{-2}$ times the error for $T_{100}$ .

Multiplying the number of subintervals, $n$ , by $10$ then gives an error that is about ${10}^{-2}$

times the error for $n$ subintervals. then conclude that

  $\left|E_{T_{10n}}\right|\approx {10}^{-2}\left|E_{T_n}\right|$

Therefore, the required pattern is $\left|E_{T_{10n}}\right|\approx {10}^{-2}\left|E_{T_n}\right|$ .

(d)

To determine

To explain: The error bound for $E_T$ accounts for the pattern.

Answer to Problem 27E

The answer is $\left|E_{T_n}\right|\le \frac{{\pi }^3}{12n^2}$

Explanation of Solution

Given information:

The integral is

Calculation:

The Error bound formula for $E_T$ is $\left|E_T\right|\le \frac{b-a}{12}{h}^2{M}_{f^{″}}$ where $a,b$ are the integrations bounds, $h$ is the length of each sub interval, $M_{f^{″}}$ is the maximum value of $\left|f^{″}\right|$ .

Hence, the integral is ${{\displaystyle \int }}_0^{\pi }\sin xdx$ then

  $\begin{array}{c}a=0,\\ b=\pi \text{,}\end{array}$

And,

  $f(x)=\sin x\cdot n$ subintervals then means

  $\begin{array}{c}h=\frac{b-a}{n}\\ =\frac{\pi -0}{n}\\ =\frac{\pi }{n}\end{array}$

Hence,

  $\begin{array}{c}f(x)=\sin x\to f^{\prime }\\ =\cos x\\ \to f^{″}(x)=-\sin x\end{array}$

And $|-\sin x|$ has a maximum value of 1, Since $-1\le \sin x\le 1$ for all $x$ then $M_{f^{″}}=1$ .

Put these into the error formula and gives:

  $\begin{array}{c}\left|E_{T_n}\right|\le \frac{b-a}{12}{h}^2{M}_{f^{″}}\\ \le \frac{\pi -0}{12}{\left(\frac{\pi }{n}\right)}^2\\ \le \frac{\pi }{12}\left(\frac{{\pi }^2}{n^2}\right)\\ \le \frac{{\pi }^3}{12n^2}\end{array}$

If the number of subintervals is multiplied by 10, then $h=\frac{\pi }{10n}$ , the error formula is then,

  $\begin{array}{c}\left|E_{T_{10n}}\right|\le \frac{b-a}{12}{h}^2{M}_{f^{″}}\\ \le \frac{\pi -0}{12}{\left(\frac{\pi }{10n}\right)}^2\\ \le \frac{\pi }{12}\left(\frac{{\pi }^2}{100n^2}\right)\\ \le \frac{{\pi }^3}{1200n^2}\\ \le \frac{1}{100}\cdot \frac{{\pi }^3}{12n^2}\\ \le {10}^{-2}\cdot \left|E_{T_n}\right|\end{array}$

Therefore, the required error bound for $E_T$ accounts for the pattern is $\left|E_{T_n}\right|\le \frac{{\pi }^3}{12n^2}$ .